1 Data

Information about the test:

question	description
z1	linear function
z2	3D
z3	arithmetic rules
z4	Easy ineq.
z5	logs
z6	logs
z7	graph translation
z8	sine pi/3
z9	trig.ineq.
z10	trig.identity
z11	period
z12	rational exponents
z13	ellipsoid
z14	limit
z15	series
z16	diff.quotient
z17	graph f'
z18	product rule
z19	quotient rule
z20	(exp)'
z21	(ln(sin))'
z22	hyp.functions
z23	slope tangent
z24	IVT
z25	velocity
z26	int(poly)
z27	int(exp)
z28	Int = 0
z29	int even funct
z31	int(abs)
z32	FtoC algebra
z33	difference vectors
z35	intersect planes
z36	parallel planes
z30	FtoC graph
z34	normal vector

Load the student scores for the test - here we load the 2017 and 2018 ETH Zurich test data:

test_scores

## # A tibble: 3,433 × 38
##    year  class    z1    z2    z3    z4    z5    z6    z7    z8    z9   z10   z11
##    <chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
##  1 2017  s21t…     1     0     1     1     1     0     1     0     1     0     1
##  2 2017  s21t…     1     0     1     1     1     1     0     1     1     2     1
##  3 2017  s21t…     1     0     0     0     1     1     1     0     1     1     1
##  4 2017  s21t…     1     0     1     1     1     1     1     1     1     1     1
##  5 2017  s21t…     1     0     1     0     2     0     1     0     1     0     2
##  6 2017  s21t…     0     1     0     0     1     2     0     2     2     2     1
##  7 2017  s21t…     1     0     1     0     2     1     0     1     2     1     0
##  8 2017  s21t…     1     1     1     1     2     1     1     2     2     2     2
##  9 2017  s21t…     1     1     0     1     1     1     1     1     1     1     1
## 10 2017  s21t…     1     0     1     0     0     1     0     0     1     1     1
## # … with 3,423 more rows, and 25 more variables: z12 <dbl>, z13 <dbl>,
## #   z14 <dbl>, z15 <dbl>, z16 <dbl>, z17 <dbl>, z18 <dbl>, z19 <dbl>,
## #   z20 <dbl>, z21 <dbl>, z22 <dbl>, z23 <dbl>, z24 <dbl>, z25 <dbl>,
## #   z26 <dbl>, z27 <dbl>, z28 <dbl>, z29 <dbl>, z30 <dbl>, z31 <dbl>,
## #   z32 <dbl>, z33 <dbl>, z34 <dbl>, z35 <dbl>, z36 <dbl>

1.1 Missing data

The data includes scores of 2 for the “I don’t know” answer option. We replace these with 0, reflecting a non-correct answer:

test_scores <- test_scores %>% 
  mutate(across(starts_with("z"), ~ ifelse(. == 2, 0, .)))

1.2 Data summary

The number of responses from each class:

test_scores <- test_scores %>% 
  rowwise() %>% 
  mutate(Total = rowSums(across(starts_with("z"))))

test_scores_summary <- test_scores %>% 
  group_by(year) %>% 
  summarise(
    n = n(),
    mean = mean(Total),
    sd = sd(Total),
    median = median(Total)
  )

test_scores_summary %>% 
  gt() %>% 
  fmt_number(columns = c("mean", "sd"), decimals = 2) %>%
  data_color(
    columns = c("n"),
    colors = scales::col_numeric(palette = c("Blues"), domain = NULL)
  )

year	n	mean	sd	median
2017	1682	18.68	8.44	19
2018	1751	19.03	8.20	19

p1 <- test_scores %>% 
  ggplot(aes(x = Total)) +
  geom_histogram(binwidth = 2) +
  #scale_x_continuous(limits = c(0,100), breaks = c(0, 50, 100)) +
  facet_grid(cols = vars(year)) +
  theme_minimal() +
  labs(x = "Total score (out of 36)",
       y = "Number of students",
       title = "ETH s21t") +
  theme(axis.title.x = element_blank(), axis.title.y = element_blank())

p2 <- test_scores_summary %>% 
  mutate(
    n = str_glue("{n}"),
    mean = str_glue("{round(mean, digits = 1)}"),
    sd = str_glue("{round(sd, digits = 1)}"),
    median = str_glue("{median}")
  ) %>% 
  pivot_longer(c(n, mean, sd, median), names_to = "layer", values_to = "label") %>% 
  mutate(layer = fct_relevel(layer, c("n", "sd", "mean", "median")) %>% fct_rev()) %>% 
  ggplot(aes(x = 80, y = layer, label = label)) +
  geom_text(size = 10 * 5/14, hjust = 1) +
  scale_x_continuous(limits = c(0,100)) +
  facet_grid(cols = vars(year)) +
  labs(y = "", x = NULL) +
  scale_y_discrete(labels = c("n" = "N", "mean" = "Mean", "median" = "Median")) +
  theme_minimal() +
  theme(axis.line = element_blank(), axis.ticks = element_blank(), axis.text.x = element_blank(),
        panel.grid = element_blank(), strip.text = element_blank())

p1 / p2 +  plot_layout(heights = c(5, 2.5))

ggsave("output/eth_pre_data-summary.pdf", units = "cm", width = 8, height = 8)

Mean and standard deviation for each item:

test_scores %>% 
  select(-class, -Total) %>% 
  group_by(year) %>% 
  skim_without_charts() %>% 
  select(-contains("character."), -contains("numeric.p"), -skim_type) %>% 
  rename(complete = complete_rate) %>% 
  # make the table wider, i.e. with separate columns for each year's results, with the year at the start of each column name
  pivot_wider(names_from = year, values_from = -c(skim_variable, year), names_glue = "{year}__{.value}") %>% 
  # put the columns in order by year
  select(sort(names(.))) %>% 
  select(skim_variable, everything()) %>% 
  # use GT to make the table look nice
  gt(rowname_col = "skim_variable") %>% 
  # group the columns from each year
  tab_spanner_delim(delim = "__") %>%
  fmt_number(columns = contains("numeric"), decimals = 2) %>%
  fmt_percent(columns = contains("complete"), decimals = 0) %>% 
  # change all the numeric.mean and numeric.sd column names to Mean and SD
  cols_label(
    .list = test_scores %>% select(year) %>% distinct() %>% transmute(col = paste0(year, "__numeric.mean"), label = "Mean") %>% deframe()
  ) %>% 
  cols_label(
    .list = test_scores %>% select(year) %>% distinct() %>% transmute(col = paste0(year, "__numeric.sd"), label = "SD") %>% deframe()
  ) %>%
  data_color(
    columns = contains("numeric.mean"),
    colors = scales::col_numeric(palette = c("Greens"), domain = NULL)
  )

	2017				2018
	complete	n_missing	Mean	SD	complete	n_missing	Mean	SD
z1	100%	0	0.95	0.22	100%	0	0.95	0.21
z2	100%	0	0.31	0.46	100%	0	0.28	0.45
z3	100%	0	0.62	0.48	100%	0	0.64	0.48
z4	100%	0	0.64	0.48	100%	0	0.62	0.48
z5	100%	0	0.47	0.50	100%	0	0.49	0.50
z6	100%	0	0.71	0.46	100%	0	0.73	0.44
z7	100%	0	0.71	0.46	100%	0	0.70	0.46
z8	100%	0	0.47	0.50	100%	0	0.50	0.50
z9	100%	0	0.46	0.50	100%	0	0.46	0.50
z10	100%	0	0.54	0.50	100%	0	0.54	0.50
z11	100%	0	0.57	0.49	100%	0	0.58	0.49
z12	100%	0	0.68	0.47	100%	0	0.71	0.45
z13	100%	0	0.37	0.48	100%	0	0.36	0.48
z14	100%	0	0.56	0.50	100%	0	0.57	0.49
z15	100%	0	0.46	0.50	100%	0	0.46	0.50
z16	100%	0	0.19	0.39	100%	0	0.20	0.40
z17	100%	0	0.59	0.49	100%	0	0.60	0.49
z18	100%	0	0.38	0.49	100%	0	0.36	0.48
z19	100%	0	0.35	0.48	100%	0	0.34	0.47
z20	100%	0	0.72	0.45	100%	0	0.72	0.45
z21	100%	0	0.51	0.50	100%	0	0.52	0.50
z22	100%	0	0.52	0.50	100%	0	0.52	0.50
z23	100%	0	0.41	0.49	100%	0	0.43	0.50
z24	100%	0	0.59	0.49	100%	0	0.58	0.49
z25	100%	0	0.54	0.50	100%	0	0.73	0.44
z26	100%	0	0.80	0.40	100%	0	0.80	0.40
z27	100%	0	0.39	0.49	100%	0	0.40	0.49
z28	100%	0	0.42	0.49	100%	0	0.46	0.50
z29	100%	0	0.48	0.50	100%	0	0.52	0.50
z30	100%	0	0.76	0.43	100%	0	0.73	0.44
z31	100%	0	0.38	0.48	100%	0	0.37	0.48
z32	100%	0	0.23	0.42	100%	0	0.20	0.40
z33	100%	0	0.78	0.42	100%	0	0.76	0.43
z34	100%	0	0.58	0.49	100%	0	0.61	0.49
z35	100%	0	0.32	0.47	100%	0	0.33	0.47
z36	100%	0	0.22	0.41	100%	0	0.23	0.42

2 Testing assumptions

Before applying IRT, we should check that the data satisfies the assumptions needed by the model. In particular, to use a 1-dimensional IRT model, we should have some evidence of unidimensionality in the test scores.

2.1 Inter-item correlations

If the test is unidimensional then we would expect student scores on pairs of items to be correlated.

This plot shows the correlations between scores on each pair of items:

item_scores <- test_scores %>% 
  select(-class, -year, -Total)

cor_ci <- psych::corCi(item_scores, plot = FALSE)

psych::cor.plot.upperLowerCi(cor_ci)

Almost all correlations are significantly different from 0 - the only one that is not is:

cor_ci$ci %>% 
  as_tibble(rownames = "corr") %>% 
  filter(p > 0.05) %>% 
  arrange(-p) %>% 
  select(-contains(".e")) %>% 
  gt() %>% 
  fmt_number(columns = 2:4, decimals = 3)

corr	lower	upper	p
z1-z2	−0.002	0.053	0.073

Thus, the overall picture is that the item scores are well correlated with each other.

2.2 Dimensionality

We use factor analysis as another way to assess unidimensionality. We are looking to see whether a 1-factor solution is feasible.

structure <- check_factorstructure(item_scores)
n <- n_factors(item_scores)

2.2.1 Is the data suitable for Factor Analysis?

KMO: The Kaiser, Meyer, Olkin (KMO) measure of sampling adequacy suggests that data seems appropriate for factor analysis (KMO = 0.95).
Sphericity: Bartlett’s test of sphericity suggests that there is sufficient significant correlation in the data for factor analysis (Chisq(630) = 30319.01, p < .001).

2.2.1.1 Method Agreement Procedure:

The choice of 6 dimensions is supported by 4 (21.05%) methods out of 19 (Optimal coordinates, Parallel analysis, Kaiser criterion, BIC).

plot(n)

summary(n) %>% gt()

n_Factors	n_Methods
1	3
2	1
3	1
4	1
6	4
8	1
11	3
33	1
34	1
35	3

#n %>% tibble() %>% gt()

fa.parallel(item_scores, fa = "fa")

## Parallel analysis suggests that the number of factors =  9  and the number of components =  NA

pdf(file = "output/eth_pre_scree.pdf", width = 6, height = 4)
fa.parallel(item_scores, fa = "fa")

## Parallel analysis suggests that the number of factors =  9  and the number of components =  NA

dev.off()

## png 
##   2

This shows that both the 6-factor and 1-factor solutions are worth exploring.

2.2.2 1 Factor

We use the factanal function to fit a 1-factor model.

Note that this function cannot handle missing data, so any NA scores must be set to 0 for this analysis.

Show factanal output

fitfact <- factanal(item_scores,
                    factors = 1,
                    rotation = "varimax")
print(fitfact, digits = 2, cutoff = 0.3, sort = TRUE)

## 
## Call:
## factanal(x = item_scores, factors = 1, rotation = "varimax")
## 
## Uniquenesses:
##   z1   z2   z3   z4   z5   z6   z7   z8   z9  z10  z11  z12  z13  z14  z15  z16 
## 0.95 0.95 0.75 0.79 0.72 0.74 0.79 0.79 0.61 0.65 0.67 0.88 0.91 0.73 0.87 0.81 
##  z17  z18  z19  z20  z21  z22  z23  z24  z25  z26  z27  z28  z29  z30  z31  z32 
## 0.78 0.78 0.78 0.63 0.59 0.61 0.64 0.73 0.94 0.69 0.63 0.64 0.85 0.81 0.78 0.78 
##  z33  z34  z35  z36 
## 0.88 0.86 0.79 0.84 
## 
## Loadings:
##  [1] 0.53 0.51 0.62 0.59 0.57 0.52 0.61 0.64 0.62 0.60 0.52 0.56 0.61 0.60     
## [16]      0.50 0.45 0.46 0.46 0.35 0.31 0.36 0.44 0.47 0.46 0.47      0.39 0.44
## [31] 0.46 0.47 0.34 0.38 0.46 0.40
## 
##                Factor1
## SS loadings       8.35
## Proportion Var    0.23
## 
## Test of the hypothesis that 1 factor is sufficient.
## The chi square statistic is 5447.17 on 594 degrees of freedom.
## The p-value is 0

load <- tidy(fitfact)

ggplot(load, aes(x = fl1, y = 0)) + 
  geom_point() + 
  geom_label_repel(aes(label = paste0("z", rownames(load))), show.legend = FALSE) +
  labs(x = "Factor 1", y = NULL,
       title = "Standardised Loadings", 
       subtitle = "Based upon correlation matrix") +
  theme_minimal()

## Warning: ggrepel: 13 unlabeled data points (too many overlaps). Consider
## increasing max.overlaps

load %>% 
  select(question = variable, factor_loading = fl1) %>% 
  left_join(item_info, by = "question") %>% 
  arrange(-factor_loading) %>% 
  gt() %>%
  data_color(
    columns = contains("factor"),
    colors = scales::col_numeric(palette = c("Greens"), domain = NULL)
  )

question	factor_loading	description
z21	0.6407106	(ln(sin))'
z9	0.6228304	trig.ineq.
z22	0.6223752	hyp.functions
z27	0.6076260	int(exp)
z20	0.6061690	(exp)'
z23	0.5998775	slope tangent
z28	0.5961137	Int = 0
z10	0.5878608	trig.identity
z11	0.5709488	period
z26	0.5590386	int(poly)
z5	0.5324571	logs
z14	0.5232019	limit
z24	0.5215894	IVT
z6	0.5086087	logs
z3	0.4972984	arithmetic rules
z19	0.4718772	quotient rule
z17	0.4705175	graph f'
z32	0.4683131	FtoC algebra
z18	0.4646006	product rule
z31	0.4643091	int(abs)
z8	0.4602708	sine pi/3
z7	0.4593494	graph translation
z35	0.4579597	intersect planes
z4	0.4547066	Easy ineq.
z30	0.4376910	FtoC graph
z16	0.4367742	diff.quotient
z36	0.4014553	parallel planes
z29	0.3875447	int even funct
z34	0.3765420	normal vector
z15	0.3603866	series
z12	0.3513586	rational exponents
z33	0.3415341	difference vectors
z13	0.3075858	ellipsoid
z25	0.2364578	velocity
z2	0.2342048	3D
z1	0.2277957	linear function

This factor appears to be dominated by “procedural calculus skills”: standard calculation of derivatives, integral and some trigonometry.

2.2.3 6 Factor

Here we also investigate the 6-factor solution, to see whether these factors are interpretable.

Show factanal output

fitfact6 <- factanal(item_scores, factors = 6, rotation = "varimax")
print(fitfact6, digits = 2, cutoff = 0.3, sort = TRUE)

## 
## Call:
## factanal(x = item_scores, factors = 6, rotation = "varimax")
## 
## Uniquenesses:
##   z1   z2   z3   z4   z5   z6   z7   z8   z9  z10  z11  z12  z13  z14  z15  z16 
## 0.91 0.81 0.74 0.75 0.68 0.74 0.72 0.76 0.53 0.64 0.64 0.87 0.80 0.71 0.82 0.72 
##  z17  z18  z19  z20  z21  z22  z23  z24  z25  z26  z27  z28  z29  z30  z31  z32 
## 0.69 0.41 0.00 0.47 0.53 0.59 0.60 0.63 0.91 0.58 0.60 0.61 0.83 0.71 0.76 0.75 
##  z33  z34  z35  z36 
## 0.84 0.70 0.56 0.70 
## 
## Loadings:
##     Factor1 Factor2 Factor3 Factor4 Factor5 Factor6
## z9   0.54    0.37                                  
## z23  0.50                                          
## z28  0.50                                          
## z24          0.52                                  
## z20                  0.59                          
## z18                          0.69                  
## z19                          0.95                  
## z35                                  0.58          
## z1                                                 
## z2                                           0.40  
## z3   0.31                                          
## z4           0.33                                  
## z5   0.38                                          
## z6                                                 
## z7           0.44                                  
## z8   0.39                                          
## z10  0.45                                          
## z11  0.43    0.36                                  
## z12                                                
## z13                                          0.34  
## z14  0.32                                          
## z15                                                
## z16  0.43                                          
## z17          0.48                                  
## z21  0.46            0.43                          
## z22  0.41            0.36                          
## z25                                                
## z26          0.34    0.48                          
## z27  0.45            0.32                          
## z29                                                
## z30          0.48                                  
## z31  0.33                                          
## z32  0.41                                          
## z33          0.33                                  
## z34                                  0.47          
## z36                                  0.43          
## 
##                Factor1 Factor2 Factor3 Factor4 Factor5 Factor6
## SS loadings        3.5    2.65    1.68    1.62    1.26    0.98
## Proportion Var     0.1    0.07    0.05    0.05    0.04    0.03
## Cumulative Var     0.1    0.17    0.22    0.26    0.30    0.32
## 
## Test of the hypothesis that 6 factors are sufficient.
## The chi square statistic is 932.36 on 429 degrees of freedom.
## The p-value is 2.36e-39

load6 <- tidy(fitfact6)

ggplot(load6, aes(x = fl1, y = fl2)) + 
  geom_point() + 
  geom_label_repel(aes(label = paste0("z", rownames(load))), show.legend = FALSE) +
  labs(x = "Factor 1", y = "Factor 2",
       title = "Standardised Loadings", 
       subtitle = "Based upon correlation matrix") +
  theme_minimal()

main_factors <- load6 %>% 
#  mutate(factorNone = 0.4) %>%  # add this to set the main factor to "None" where all loadings are below 0.4
  pivot_longer(names_to = "factor",
               cols = contains("fl")) %>% 
  mutate(value_abs = abs(value)) %>% 
  group_by(variable) %>% 
  top_n(1, value_abs) %>% 
  ungroup() %>% 
  transmute(main_factor = factor, variable)

load6 %>% 
  select(-uniqueness) %>% 
  # add the info about which is the main factor
  left_join(main_factors, by = "variable") %>%
  left_join(item_info %>% select(variable = question, description), by = "variable") %>% 
  arrange(main_factor) %>% 
  select(main_factor, everything()) %>% 
  # arrange adjectives by descending loading on main factor
  rowwise() %>% 
  mutate(max_loading = max(abs(c_across(starts_with("fl"))))) %>% 
  group_by(main_factor) %>% 
  arrange(-max_loading, .by_group = TRUE) %>% 
  select(-max_loading) %>% 
  # sort out the presentation
  rename("Main Factor" = main_factor, # the _ throws a latex error
         "Question" = variable) %>%
  mutate_at(
    vars(starts_with("fl")),
    ~ cell_spec(round(., digits = 3), bold = if_else(abs(.) > 0.4, T, F))
  ) %>% 
  kable(booktabs = T, escape = F, longtable = T) %>% 
  kableExtra::collapse_rows(columns = 1, valign = "top") %>%
  kableExtra::kable_styling(latex_options = c("repeat_header"))

Main Factor	Question	fl1	fl2	fl3	fl4	fl5	fl6	description
fl1	z9	0.543	0.371	0.045	0.1	0.13	0.086	trig.ineq.
fl1	z23	0.504	0.289	0.189	0.085	0.136	0.023	slope tangent
fl1	z28	0.503	0.292	0.145	0.094	0.147	0.048	Int = 0
fl1	z21	0.456	0.194	0.429	0.13	0.143	0.07	(ln(sin))’
fl1	z27	0.453	0.193	0.316	0.086	0.216	0.075	int(exp)
fl1	z10	0.447	0.239	0.234	0.104	0.084	0.182	trig.identity
fl1	z16	0.427	-0.017	0.141	0.129	0.128	0.217	diff.quotient
fl1	z11	0.426	0.357	0.127	0.065	0.082	0.15	period
fl1	z22	0.408	0.237	0.36	0.108	0.21	0.073	hyp.functions
fl1	z32	0.407	0.122	0.106	0.104	0.19	0.115	FtoC algebra
fl1	z8	0.386	0.145	0.138	0.12	0.058	0.186	sine pi/3
fl1	z5	0.379	0.182	0.2	0.093	0.114	0.278	logs
fl1	z31	0.325	0.262	0.059	0.07	0.178	0.151	int(abs)
fl1	z14	0.324	0.176	0.266	0.118	0.18	0.18	limit
fl1	z3	0.313	0.212	0.247	0.085	0.08	0.213	arithmetic rules
fl1	z6	0.293	0.268	0.257	0.072	0.122	0.136	logs
fl2	z24	0.272	0.516	0.074	0.079	0.134	0.049	IVT
fl2	z30	0.171	0.484	0.086	0.051	0.094	0.08	FtoC graph
fl2	z17	0.193	0.478	0.118	0.073	0.061	0.137	graph f’
fl2	z7	0.167	0.443	0.157	0.036	0.095	0.161	graph translation
fl2	z4	0.225	0.334	0.118	0.064	0.084	0.249	Easy ineq.
fl2	z33	0.101	0.328	0.089	0.036	0.113	0.158	difference vectors
fl2	z1	0.046	0.27	0.089	0.027	0.029	0.06	linear function
fl2	z25	0.056	0.264	0.099	0.009	0.055	0.054	velocity
fl2	z12	0.185	0.191	0.143	0.083	0.072	0.168	rational exponents
fl3	z20	0.286	0.268	0.593	0.09	0.115	0.059	(exp)’
fl3	z26	0.221	0.338	0.476	0.11	0.142	0.002	int(poly)
fl3	z29	0.23	0.126	0.27	0.093	0.145	0.03	int even funct
fl4	z19	0.227	0.071	0.129	0.951	0.101	0.085	quotient rule
fl4	z18	0.211	0.138	0.135	0.692	0.114	0.106	product rule
fl5	z35	0.216	0.133	0.114	0.104	0.583	0.102	intersect planes
fl5	z34	0.109	0.185	0.169	0.057	0.47	0.006	normal vector
fl5	z36	0.223	0.106	0.06	0.045	0.43	0.212	parallel planes
fl6	z2	0.093	0.117	0.011	0.046	0.024	0.403	3D
fl6	z13	0.099	0.255	-0.006	0.044	0.106	0.339	ellipsoid
fl6	z15	0.154	0.233	0.084	0.023	0.143	0.28	series

Here, the first factor seems better described by “abstract understanding of calculus”. The second is based on “graphical understanding of calculus”.

The third and fourth factors are “procedural calculus skills”, with the product and quotient rules distinguished in the fourth factor.

The fifth and sixth factors have a geometrical flavour, with the fifth factor in particular being about “vector geometry”.

2.2.4 Conclusion

The 6-factor solution does seem to be interpretable, however the 1-factor solution is still reasonable. In terms of explaining the variance in the data, the 1-factor solution does a good job at capturing a large chunk:

# Compute the proportion of variance explained; see https://stackoverflow.com/a/58159992/17459126
x1 <- loadings(fitfact)
propvar1 <- colSums(x1^2)/nrow(x1)

rbind(`Proportion Var` = propvar1, `Cumulative Var` = cumsum(propvar1))

##                  Factor1
## Proportion Var 0.2320524
## Cumulative Var 0.2320524

x6 <- loadings(fitfact6)
propvar6 <- colSums(x6^2)/nrow(x6)

rbind(`Proportion Var` = propvar6, `Cumulative Var` = cumsum(propvar6))

##                   Factor1    Factor2    Factor3    Factor4    Factor5
## Proportion Var 0.09717865 0.07370191 0.04679681 0.04500204 0.03500279
## Cumulative Var 0.09717865 0.17088056 0.21767737 0.26267941 0.29768220
##                   Factor6
## Proportion Var 0.02729802
## Cumulative Var 0.32498023

Therefore, we are justified in proceeding with a unidimensional IRT model.

3 Fitting 2 parameter logistic MIRT model

We can fit a Multidimensional Item Response Theory (mirt) model. From the function definition:

mirt fits a maximum likelihood (or maximum a posteriori) factor analysis model to any mixture of dichotomous and polytomous data under the item response theory paradigm using either Cai's (2010) Metropolis-Hastings Robbins-Monro (MHRM) algorithm.

The process is to first fit the model, and save the result as a model object that we can then parse to get tabular or visual displays of the model that we might want. When fitting the model, we have the option to specify a few arguments, which then get interpreted as parameters to be passed to the model.

Show model fitting output

fit_2pl <- mirt(
  data = item_scores, # just the columns with question scores
  model = 1,          # number of factors to extract
  itemtype = "2PL",   # 2 parameter logistic model
  SE = TRUE           # estimate standard errors
  )

## 
Iteration: 1, Log-Lik: -66668.278, Max-Change: 0.71582
Iteration: 2, Log-Lik: -65483.178, Max-Change: 0.26188
Iteration: 3, Log-Lik: -65316.598, Max-Change: 0.14767
Iteration: 4, Log-Lik: -65249.858, Max-Change: 0.11072
Iteration: 5, Log-Lik: -65212.345, Max-Change: 0.09658
Iteration: 6, Log-Lik: -65190.906, Max-Change: 0.06270
Iteration: 7, Log-Lik: -65178.102, Max-Change: 0.06095
Iteration: 8, Log-Lik: -65169.976, Max-Change: 0.03668
Iteration: 9, Log-Lik: -65165.338, Max-Change: 0.03883
Iteration: 10, Log-Lik: -65162.425, Max-Change: 0.02443
Iteration: 11, Log-Lik: -65160.764, Max-Change: 0.01661
Iteration: 12, Log-Lik: -65159.750, Max-Change: 0.01247
Iteration: 13, Log-Lik: -65158.047, Max-Change: 0.00237
Iteration: 14, Log-Lik: -65158.024, Max-Change: 0.00195
Iteration: 15, Log-Lik: -65158.012, Max-Change: 0.00168
Iteration: 16, Log-Lik: -65157.995, Max-Change: 0.00082
Iteration: 17, Log-Lik: -65157.990, Max-Change: 0.00089
Iteration: 18, Log-Lik: -65157.986, Max-Change: 0.00072
Iteration: 19, Log-Lik: -65157.976, Max-Change: 0.00018
Iteration: 20, Log-Lik: -65157.975, Max-Change: 0.00020
Iteration: 21, Log-Lik: -65157.975, Max-Change: 0.00019
Iteration: 22, Log-Lik: -65157.974, Max-Change: 0.00012
Iteration: 23, Log-Lik: -65157.973, Max-Change: 0.00012
Iteration: 24, Log-Lik: -65157.973, Max-Change: 0.00011
Iteration: 25, Log-Lik: -65157.973, Max-Change: 0.00009
## 
## Calculating information matrix...

3.1 Local independence

We compute Yen’s \(Q_3\) (1984) to check for any dependence between items after controlling for \(\theta\). This gives a score for each pair of items, with scores above 0.2 regarded as problematic (see DeMars, p. 48).

residuals_2pl  %>% as.matrix() %>% 
  corrplot::corrplot(type = "upper")

This shows that most item pairs are independent, with only one pair showing cause for concern:

residuals_2pl %>%
  rownames_to_column(var = "q1") %>%
  as_tibble() %>% 
  pivot_longer(cols = starts_with("z"), names_to = "q2", values_to = "Q3_score") %>% 
  filter(abs(Q3_score) > 0.2) %>% 
  filter(parse_number(q1) < parse_number(q2)) %>%
  gt()

q1	q2	Q3_score
z18	z19	0.6740812
z34	z35	0.2109123

Items z18 and z19 are on the product and quotient rules.

Given that this violation of the local independence assumption is very mild, we proceed using this model.

3.2 Model parameters

We then compute factor score estimates and augment the existing data frame with these estimates, to keep everything in one place. To do the estimation, we use the fscores() function from the mirt package which takes in a computed model object and computes factor score estimates according to the method specified. We will use the EAP method for factor score estimation, which is the “expected a-posteriori” method, the default. We specify it explicitly below, but the results would have been the same if we omitted specifying the method argument since it’s the default method the function uses.

test_scores_with_2pl_ability <- test_scores %>%
  bind_cols(fscores(fit_2pl, method = "EAP"))

We can also calculate the model coefficient estimates using a generic function coef() which is used to extract model coefficients from objects returned by modeling functions. We will set the IRTpars argument to TRUE, which means slope intercept parameters will be converted into traditional IRT parameters.

coefs_2pl <- coef(fit_2pl, IRTpars = TRUE)

The resulting object coefs is a list, with one element for each question, and an additional GroupPars element that we won’t be using. The output is a bit long, so we’re only showing a few of the elements here:

coefs_2pl[1:3]

## $z1
##                a         b  g  u
## par     1.219104 -2.946041  0  1
## CI_2.5  1.024917 -3.298279 NA NA
## CI_97.5 1.413291 -2.593803 NA NA
## 
## $z2
##                 a        b  g  u
## par     0.5612745 1.670103  0  1
## CI_2.5  0.4752132 1.405353 NA NA
## CI_97.5 0.6473358 1.934852 NA NA
## 
## $z3
##                a          b  g  u
## par     1.354723 -0.5385656  0  1
## CI_2.5  1.236060 -0.6126183 NA NA
## CI_97.5 1.473386 -0.4645129 NA NA

# coefs_2pl[35:37]

Let’s take a closer look at the first element:

coefs_2pl[1]

## $z1
##                a         b  g  u
## par     1.219104 -2.946041  0  1
## CI_2.5  1.024917 -3.298279 NA NA
## CI_97.5 1.413291 -2.593803 NA NA

In this output:

a is discrimination
b is difficulty
endpoints of the 95% confidence intervals are also shown

To make this output more user friendly, we have produced a function tidy_mirt_coeffs that transforms it into a tidy table.

source("fn_tidy_mirt_coefs.R")

tidy_2pl <- tidy_mirt_coefs(coefs_2pl) %>% 
  # keep only the a and b parameters from the 2PL model
  filter(par %in% c("a", "b"))

Here is the result:

tidy_2pl %>%
  mutate(ci = str_glue("[{round(CI_2.5, 2)}, {round(CI_97.5, 2)}]")) %>% 
  select(-starts_with("CI_")) %>% 
  pivot_wider(names_from = "par", values_from = c(est, ci), names_glue = "{par}_{.value}") %>% 
  gt() %>% 
  fmt_number(columns = contains("_est"), decimals = 2) %>%
  data_color(
    columns = contains("a_est"),
    colors = scales::col_numeric(palette = c("Greens"), domain = NULL)
  ) %>%
  data_color(
    columns = contains("b_est"),
    colors = scales::col_numeric(palette = c("Blues"), domain = NULL)
  ) %>%
  tab_spanner(label = "Discrimination", columns = contains("a_")) %>%
  tab_spanner(label = "Difficulty", columns = contains("b_")) %>%
  cols_label(
    a_est = "Est.",
    b_est = "Est.",
    a_ci = "95% CI",
    b_ci = "95% CI"
  )

Question	Discrimination		Difficulty
Question	Est.	95% CI	Est.	95% CI
z1	1.22	[1.02, 1.41]	−2.95	[-3.3, -2.59]
z2	0.56	[0.48, 0.65]	1.67	[1.41, 1.93]
z3	1.35	[1.24, 1.47]	−0.54	[-0.61, -0.46]
z4	1.18	[1.07, 1.29]	−0.59	[-0.67, -0.5]
z5	1.47	[1.34, 1.59]	0.08	[0.02, 0.14]
z6	1.56	[1.42, 1.7]	−0.86	[-0.94, -0.78]
z7	1.29	[1.17, 1.41]	−0.88	[-0.97, -0.8]
z8	1.16	[1.06, 1.27]	0.07	[-0.01, 0.14]
z9	2.02	[1.86, 2.18]	0.13	[0.07, 0.18]
z10	1.77	[1.63, 1.91]	−0.14	[-0.2, -0.09]
z11	1.68	[1.55, 1.82]	−0.28	[-0.34, -0.22]
z12	0.89	[0.79, 0.99]	−1.09	[-1.22, -0.96]
z13	0.72	[0.63, 0.81]	0.85	[0.72, 0.99]
z14	1.44	[1.31, 1.56]	−0.26	[-0.32, -0.19]
z15	0.85	[0.76, 0.94]	0.21	[0.12, 0.3]
z16	1.54	[1.39, 1.69]	1.30	[1.19, 1.4]
z17	1.23	[1.12, 1.34]	−0.40	[-0.48, -0.33]
z18	1.24	[1.12, 1.35]	0.55	[0.47, 0.63]
z19	1.31	[1.19, 1.43]	0.64	[0.56, 0.72]
z20	2.33	[2.13, 2.53]	−0.74	[-0.8, -0.68]
z21	2.14	[1.97, 2.31]	−0.05	[-0.1, 0.01]
z22	2.00	[1.84, 2.16]	−0.05	[-0.11, 0]
z23	1.88	[1.72, 2.03]	0.27	[0.21, 0.33]
z24	1.44	[1.32, 1.56]	−0.31	[-0.38, -0.25]
z25	0.53	[0.45, 0.62]	−1.13	[-1.33, -0.92]
z26	2.42	[2.2, 2.65]	−1.04	[-1.1, -0.97]
z27	1.99	[1.83, 2.16]	0.35	[0.29, 0.41]
z28	1.85	[1.7, 2]	0.21	[0.15, 0.27]
z29	0.91	[0.82, 1.01]	0.00	[-0.09, 0.09]
z30	1.26	[1.14, 1.38]	−1.08	[-1.19, -0.98]
z31	1.25	[1.14, 1.36]	0.54	[0.46, 0.62]
z32	1.60	[1.45, 1.75]	1.15	[1.06, 1.24]
z33	0.94	[0.83, 1.04]	−1.51	[-1.67, -1.35]
z34	0.91	[0.81, 1]	−0.51	[-0.61, -0.42]
z35	1.28	[1.16, 1.4]	0.73	[0.65, 0.82]
z36	1.25	[1.12, 1.37]	1.28	[1.16, 1.39]

tidy_2pl_wide <- tidy_2pl %>% 
  pivot_wider(names_from = "par", values_from = c(est, CI_2.5, CI_97.5), names_glue = "{par}_{.value}")

wright_plot <- tidy_2pl_wide %>% 
  mutate(qnum = parse_number(Question)) %>% 
  ggplot(aes(
    x = a_est,
    y = b_est
  )) +
  geom_errorbar(aes(ymin = b_CI_2.5, ymax = b_CI_97.5), width = 0, alpha = 0.5) +
  geom_errorbar(aes(xmin = a_CI_2.5, xmax = a_CI_97.5), width = 0, alpha = 0.5) +
  geom_text_repel(aes(label = Question), alpha = 0.5) +
  geom_point() +
  theme_minimal() +
  labs(x = "Discrimination",
       y = "Difficulty")

wright_plot

tidy_2pl_wide %>% 
  write_csv("output/eth_pre_2pl-results.csv")

3.3 Comparing years and classes

Do students from different programmes of study have different distributions of ability?

3.3.1 Differences between years

Compare the distribution of abilities in the year groups (though in this case there is only one).

ggplot(test_scores_with_2pl_ability, aes(F1, fill = as.factor(year), colour = as.factor(year))) +
  geom_density(alpha=0.5) + 
  scale_x_continuous(limits = c(-3.5,3.5)) +
  labs(title = "Density plot", 
       subtitle = "Ability grouped by year of taking the test", 
       x = "Ability", y = "Density",
       fill = "Year", colour = "Year") +
  theme_minimal()

3.3.2 Differences between classes

Compare the distribution of abilities in the various classes.

ggplot(test_scores_with_2pl_ability, aes(x = F1, y = class, colour = class, fill = class)) +
  geom_density_ridges(alpha = 0.5) + 
  scale_x_continuous(limits = c(-3.5,3.5)) +
  guides(fill = FALSE, colour = FALSE) +
  labs(title = "Density plot", 
       subtitle = "Ability grouped by class of taking the test", 
       x = "Ability", y = "Class") +
  theme_minimal()

## Warning: `guides(<scale> = FALSE)` is deprecated. Please use `guides(<scale> =
## "none")` instead.

## Picking joint bandwidth of 0.194

3.4 Information curves

theta <- seq(-6, 6, by=0.05)

info_matrix <- testinfo(fit_2pl, theta, individual = TRUE)
colnames(info_matrix) <- item_info %>% pull(question)
item_info_data <- info_matrix %>% 
  as_tibble() %>% 
  bind_cols(theta = theta) %>% 
  pivot_longer(cols = -theta, names_to = "item", values_to = "info_y") %>% 
  left_join(item_info %>% select(item = question), by = "item") %>% 
  mutate(item = fct_reorder(item, parse_number(item)))

3.4.1 Test information curve

item_info_data %>% 
  group_by(theta) %>% 
  summarise(info_y = sum(info_y)) %>% 
  ggplot(aes(x = theta, y = info_y)) +
  geom_line() +
  labs(x = "Ability", y = "Information", title = "ETH s21t") +
  theme_minimal()

ggsave("output/eth_pre_info.pdf", width = 10, height = 6, units = "cm")

This shows that the information given by the test is focused reasonably centrally around the mean ability level.

3.4.2 Item information curves

Breaking this down by question helps to highlight those questions that are most/least informative:

item_info_data %>% 
  ggplot(aes(x = theta, y = info_y, colour = item)) +
  geom_line() +
  scale_colour_viridis_d(option = "plasma", end = 0.8, direction = -1) +
  facet_wrap(vars(item)) +
  labs(y = "Information") +
  theme_minimal()

3.5 Total information

Using mirt’s areainfo function, we can find the total area under the information curves.

irt_info <- areainfo(fit_2pl, c(-4,4))
irt_info %>% gt()

LowerBound	UpperBound	Info	TotalInfo	Proportion	nitems
-4	4	49.13213	50.42239	0.9744109	36

This shows that the total information in all items is 50.4223874.

3.5.1 Information by item

tidy_info <- item_info %>%
  mutate(item_num = row_number()) %>% 
  mutate(TotalInfo = purrr::map_dbl(
    item_num,
    ~ areainfo(fit_2pl,
               c(-4, 4),
               which.items = .x) %>% pull(TotalInfo)
  ))

tidy_info %>%
  select(-item_num) %>% 
  arrange(-TotalInfo) %>% 
  #group_by(outcome) %>% 
  gt() %>% 
  fmt_number(columns = contains("a_"), decimals = 2) %>%
  fmt_number(columns = contains("b_"), decimals = 2) %>%
  data_color(
    columns = contains("info"),
    colors = scales::col_numeric(palette = c("Greens"), domain = NULL)
  ) %>%
  data_color(
    columns = contains("outcome"),
    colors = scales::col_factor(palette = c("viridis"), domain = NULL)
  ) %>%
  cols_label(
    TotalInfo = "Information"
  )

question	description	Information
z26	int(poly)	2.4238253
z20	(exp)'	2.3258216
z21	(ln(sin))'	2.1365919
z9	trig.ineq.	2.0158405
z22	hyp.functions	1.9989907
z27	int(exp)	1.9938718
z23	slope tangent	1.8750694
z28	Int = 0	1.8501832
z10	trig.identity	1.7675098
z11	period	1.6839435
z33	difference vectors	1.6041930
z6	logs	1.5577381
z16	diff.quotient	1.5386847
z5	logs	1.4663312
z24	IVT	1.4390034
z14	limit	1.4350697
z3	arithmetic rules	1.3547187
z19	quotient rule	1.3093576
z7	graph translation	1.2867712
z30	FtoC graph	1.2809342
z31	int(abs)	1.2631005
z32	FtoC algebra	1.2503874
z34	normal vector	1.2462025
z18	product rule	1.2373800
z17	graph f'	1.2268893
z1	linear function	1.2188793
z4	Easy ineq.	1.1804333
z8	sine pi/3	1.1623303
z35	intersect planes	0.9353326
z29	int even funct	0.9128694
z36	parallel planes	0.9058856
z12	rational exponents	0.8887859
z15	series	0.8494321
z13	ellipsoid	0.7169794
z2	3D	0.5552891
z25	velocity	0.5277613

Restricting instead to the range \(-2\leq\theta\leq2\):

tidy_info <- item_info %>%
  mutate(item_num = row_number()) %>% 
  mutate(TotalInfo = purrr::map_dbl(
    item_num,
    ~ areainfo(fit_2pl,
               c(-2, 2),
               which.items = .x) %>% pull(Info)
  ))

tidy_info %>%
  select(-item_num) %>% 
  arrange(-TotalInfo) %>% 
  #group_by(outcome) %>% 
  gt() %>% 
  fmt_number(columns = contains("a_"), decimals = 2) %>%
  fmt_number(columns = contains("b_"), decimals = 2) %>%
  data_color(
    columns = contains("info"),
    colors = scales::col_numeric(palette = c("Greens"), domain = NULL)
  ) %>%
  data_color(
    columns = contains("outcome"),
    colors = scales::col_factor(palette = c("viridis"), domain = NULL)
  ) %>%
  cols_label(
    TotalInfo = "Information"
  )

question	description	Information
z26	int(poly)	2.2090139
z20	(exp)'	2.2045246
z21	(ln(sin))'	2.0775693
z9	trig.ineq.	1.9433205
z22	hyp.functions	1.9265566
z27	int(exp)	1.9039799
z23	slope tangent	1.7785873
z28	Int = 0	1.7545283
z10	trig.identity	1.6643308
z11	period	1.5605291
z5	logs	1.3171858
z6	logs	1.3144395
z24	IVT	1.2724263
z14	limit	1.2722619
z33	difference vectors	1.2662362
z3	arithmetic rules	1.1482188
z16	diff.quotient	1.1403300
z19	quotient rule	1.0806177
z30	FtoC graph	1.0322320
z32	FtoC algebra	1.0269435
z17	graph f'	1.0141086
z18	product rule	1.0104220
z7	graph translation	1.0086145
z8	sine pi/3	0.9547704
z4	Easy ineq.	0.9399399
z31	int(abs)	0.9357119
z34	normal vector	0.8652332
z29	int even funct	0.6597833
z36	parallel planes	0.6349100
z15	series	0.5847449
z12	rational exponents	0.5619252
z35	intersect planes	0.5401834
z13	ellipsoid	0.4171787
z1	linear function	0.2895158
z25	velocity	0.2433460
z2	3D	0.2430917

3.6 Response curves

These show the probability of a correct response at each ability level:

trace_data <- probtrace(fit_2pl, theta) %>% 
  as_tibble() %>% 
  bind_cols(theta = theta) %>% 
  pivot_longer(cols = -theta, names_to = "level", values_to = "y") %>% 
  separate(level, into = c("item", NA, "score"), sep = "\\.") %>% 
  mutate(item = fct_reorder(item, parse_number(item))) %>% 
  filter(score == 1)

trace_data %>% 
  ggplot(aes(x = theta, y = y, colour = score)) +
  geom_line() +
  scale_colour_viridis_d(option = "plasma", end = 0.8, direction = -1, guide = "none") +
  facet_wrap(vars(item)) +
  labs(y = "Probability of correct response") +
  theme_minimal()

These can all be overlaid on a single plot:

plt <- trace_data %>% 
  ggplot(aes(x = theta, y = y, colour = item, text = item)) +
  geom_line() +
  scale_colour_viridis_d(option = "plasma", end = 0.8, direction = -1) +
  labs(y = "Probability of correct response") +
  theme_minimal()

ggplotly(plt, tooltip = "text")

ggsave(plot = plt, file = "output/eth_pre_iccs-superimposed.pdf", width = 20, height = 14, units = "cm")

Here we highlight some items to illustrate the role of the difficulty and discrimination parameters

highlight_items <- c("z1", "z2", "z26", "z27")
item_colours <- c(viridis::viridis_pal(option = "plasma", end = 0.8, direction = -1)(length(highlight_items)), "black")

wright_plot_shaded <- tidy_2pl_wide %>%
  mutate(item = Question) %>%
  mutate(
    item_colour = case_when(item %in% highlight_items ~ item,
                            TRUE ~ "ignore"),
    item_size = if_else(item %in% highlight_items, 4, 3)
  ) %>% 
  ggplot(aes(
    x = a_est,
    y = b_est,
    colour = item_colour
  )) +
  geom_errorbar(aes(ymin = b_CI_2.5, ymax = b_CI_97.5), width = 0, alpha = 0.5) +
  geom_errorbar(aes(xmin = a_CI_2.5, xmax = a_CI_97.5), width = 0, alpha = 0.5) +
  geom_text_repel(aes(label = Question, size = item_size), alpha = 0.5) +
  geom_point() +
  scale_colour_manual("item", values = item_colours, limits = highlight_items, guide = "none") +
  scale_size(guide = "none", range = c(2.5, 4.5)) +
  theme_minimal() +
  coord_flip() +
  labs(x = "Discrimination",
       y = "Difficulty")

trace_examples <- trace_data %>% 
  filter(item %in% highlight_items) %>% 
  ggplot(aes(x = theta, y = y, colour = item)) +
  geom_line() +
  # ggrepel::geom_label_repel(
  #   data = trace_data %>% filter(item %in% highlight_items, y > 0.5) %>% group_by(item) %>% slice_min(y, n = 1),
  #   aes(label = item),
  #   box.padding = 0.5,
  #   min.segment.length = Inf,
  #   show.legend = FALSE
  # ) +
  geom_label(
    data = trace_data %>% filter(item %in% c("z1", "z27"), y > 0.45) %>% group_by(item) %>% slice_min(y, n = 1),
    aes(label = item),
    show.legend = FALSE
  ) +
  geom_label(
    data = trace_data %>% filter(item %in% c("z2", "z26"), y > 0.55) %>% group_by(item) %>% slice_min(y, n = 1),
    aes(label = item),
    show.legend = FALSE
  ) +
  scale_colour_viridis_d("item", option = "plasma", end = 0.8, direction = -1, guide = "none") +
  labs(x = "Ability", y = "Probability of correct response") +
  theme_minimal()

combined_plot <- patchwork::wrap_plots(wright_plot_shaded, trace_examples)
ggsave("output/eth_pre_wrightmap_highlight.pdf", plot = combined_plot, units = "cm", width = 18, height = 9)
combined_plot

removed_items <- read_csv("data-eth/eth-metadata.csv") %>% filter(is.na(post)) %>% mutate(pre = str_replace(pre, "A", "z")) %>% select(pre) %>% deframe()
highlight_removed_items <- trace_data %>% 
  mutate(highlight_item = item %in% removed_items) %>% 
  mutate(line_width = ifelse(highlight_item, 1, 0.5))

highlight_removed_items %>% 
  ggplot(aes(x = theta, y = y, colour = item, text = item, alpha = highlight_item)) +
  geom_line(aes(size = highlight_item)) +
  #geom_point(data = highlight_removed_items %>% filter(highlight_item == TRUE, theta == 0)) +
  ggrepel::geom_label_repel(
    #data = highlight_removed_items %>% filter(highlight_item == TRUE) %>% group_by(item) %>% mutate(item_num = cur_group_id()) %>% ungroup() %>% mutate(threshold = seq(5, -5, len=max(item_num))[item_num]) %>% filter(theta >= threshold) %>% group_by(item_num) %>% slice(n=1),
    #data = highlight_removed_items %>% filter(highlight_item == TRUE, y >= 0.5) %>% group_by(item) %>% slice_min(y, n = 1),
    data = highlight_removed_items %>% filter(highlight_item == TRUE, theta == 0),
    aes(label = item),
    box.padding = 0.5,
    max.overlaps = Inf,
    min.segment.length = 0,
    show.legend = FALSE
  ) +
  scale_colour_viridis_d("Question", option = "plasma", end = 0.8, direction = -1) +
  scale_size_manual(values = c("FALSE" = 0.6, "TRUE" = 0.9), guide = "none") +
  scale_alpha_discrete(guide = "none", range = c(0.2, 1)) +
  labs(x = "Ability", y = "Expected score") +
  theme_minimal() +
  theme(legend.position="bottom",#legend.title=element_blank(),
      legend.margin = margin(0, 0, 0, 0),
      legend.spacing.x = unit(1, "mm"),
      legend.spacing.y = unit(0, "mm")) +
  guides(colour = guide_legend(nrow = 4))

ggsave(file = "output/eth_pre_iccs-highlight.pdf", width = 16, height = 12, units = "cm")

4 Checking 3PL

The 3 parameter IRT model adds a third “guessing” parameter for each item, which gives the lower asymptote of the response curve.

Given that the test offered an “I don’t know” option (which we have mapped to a score of 0), we would expect that students are less likely to be guessing when answering these multiple-choice questions. So, we would expect to see the lower aymptotes being closer to 0 than would ordinarily be expected for a 4- or 5-option MCQ.

fit_3pl <- mirt(
  data = item_scores, # just the columns with question scores
  model = 1,          # number of factors to extract
  itemtype = "3PL",   # 3 parameter logistic model
  SE = TRUE           # estimate standard errors
  )

residuals_3pl <- residuals(fit_3pl, type = "Q3") %>% data.frame

coefs_3pl <- coef(fit_3pl, IRTpars = TRUE)

test_scores_with_3pl_ability <- test_scores %>%
  bind_cols(fscores(fit_3pl, method = "EAP"))

4.1 Local independence

residuals_3pl  %>% as.matrix() %>% 
  corrplot::corrplot(type = "upper")

This shows that most item pairs are independent, with only a couple of pairs showing cause for concern:

residuals_3pl %>%
  rownames_to_column(var = "q1") %>%
  as_tibble() %>% 
  pivot_longer(cols = starts_with("z"), names_to = "q2", values_to = "Q3_score") %>% 
  filter(abs(Q3_score) > 0.2) %>% 
  filter(parse_number(q1) < parse_number(q2)) %>%
  gt()

q1	q2	Q3_score
z18	z19	0.6745194
z34	z35	0.2115169

Items A18 and A19 are on the product and quotient rules, with an identical structure in each case (giving all the values that need to be used in the relevant formula).
Items A34 and A35 are are both about planes in vector geometry.